Question

If $a$ and $b$ are real numbers between $0$ and $1$ such that the points $z_1=a+i,z_2=1+bi$ and $z_3=0$ form an equilateral triangle, then $a$ and $b$ are

• A. $2+\sqrt{3},2-\sqrt{3}$
• B. $2-\sqrt{3},2-\sqrt{3}$
• C. $2-\sqrt{3},2+\sqrt{3}$
• D. None of these

Answer 1

Answer: (B)

$2-\sqrt{3},2-\sqrt{3}$

We know that the triangle with vertices $z_1,z_2,z_3$ is an equilateral if

${z_1}^2+{z_2}^2+{z_3}^3=z_1{z}_2+z_2{z}_3+z_3{z}_1$

$\therefore$ The triangle with vertices $z_1=a+i,z_2=1+bi$ and $z_3=0$ will be equilateral if

${(a-i)}^2+{(1+bi)}^2+0=(a+i)(1+bi)+0+0$

$\Rightarrow a^2-1+2ai+1-b^2+2bi=(a-b)+i(1+ab)$

$\Rightarrow a^2-b^2=a-b$ ...(1)

and $2(a+b)=1+ab$ ....(2)

(equating real and imaginary part)

Equation (1) $\Rightarrow (a-b)(a+b-1)=0\Rightarrow a=b$ or $a=1-b$

Substituting the value of $a-b$ in (2), we get

$2(a+a)=1+a^2\Rightarrow a^2-4a+1=0\Rightarrow a=\frac{4\pm \sqrt{16-4}}{2}=2\pm \sqrt{3}$

Since $0<a<1$ and $0<b<1$

$\therefore a=b=2-\sqrt{3}$

Substituting $a+b=1$ in (2), we get

$2=1+a(1-a)\Rightarrow a^2-a+1=0$

which gives imaginary value of $a$ and $b$

Hence, $a=b=2-\sqrt{3}$